Probability Uniform Distribution

Hi, welcome back to www.educator.com lectures on probability, my name is Will Murray.0000

Today, we are going to talk about the first of several continuous distributions.0004

This is probably the easiest one, it is called the uniform distribution.0010

We will see very soon why it is called that.0013

Let us jump right on in.0016

The idea of the uniform distribution is you have a finite range from θ1 to θ2.0017

Let me go ahead and draw a graph of this, as I'm talking about this.0023

You have 2 constant values, here is θ1 and then you have θ2, somewhere a bit bigger than θ1.0028

And then, you just divide your density evenly, you distribute it evenly over that range0037

which means you just take a completely horizontal line over that range.0044

What that means is, remember the total density always has to be 1,0051

that the total area under a density function always has to be 1.0055

In order to have that area be 1, the width is θ2 – θ1, the height has to be the constant 1/θ2 – θ1.0060

By the way, this triple equal sign, that means always equals 2.0070

It means that the density function is constant.0079

That is much different from all of the other density functions that we will be studying later.0082

That is what makes the uniform distribution a lot easier than some of the later ones.0087

It is that the density function is always equal to a constant.0091

That constant has to be 111, in order to give your total area 1.0094

Each part of the region is equally possible or equally probable.0101

It is very easy to calculate probabilities with the uniform distribution, if you have two values A and B here.0106

Let me go ahead and draw them in on my graph A and B.0114

If you have two values A and B, and we want to find the probability that your random variable0118

land somewhere between A and B.0124

It is very simple, you just have to calculate the distance between A and B.0126

Essentially, you are calculating this area right here.0131

And that black area is just going to be B – A/θ2 – tθ1, because it has width B - A and it has height of 111.0136

It is very easy to calculate probabilities using uniform distribution.0149

You just look at the two ranges that you are interested in, subtract them,0153

and then divide that by the appropriate constant which is always θ2 – θ1.0157

Let us see how that plays out.0164

The key properties in uniform distribution, they mean should be kind of intuitively obvious.0167

Let me draw again the graph of the uniform distribution.0173

There is θ1 and there is θ2.0178

Remember that, we are distributing the density completely evenly between θ1 and θ2.0181

You would expect the mean to be halfway between them.0187

In fact, that is where it turns out to be.0191

The mean is exactly the average of θ1 and θ2.0193

You just get θ1 + θ2/2 for the μ.0198

That is really intuitively clear, it should not be hard to remember because it should be obvious.0202

The variance is less obvious, the variance turns out to be θ2 – θ1²/12.0207

I think that is not something that you would guess.0215

You probably would guess the mean if you give a little bit of thought.0218

The variance is not something that you probably guessed.0221

You probably have to calculate it out or just memorize it.0225

The standard deviation, remember, it is always the square root of the variance.0229

If you take the square root of the variance here, you get θ2 – θ1 ÷ √ 12 which is 2 × √ 3.0233

That is the standard deviation of the uniform distribution.0243

It should still make a rough intuitive sense because it is really a measure of how spread out the interval is.0249

Remember that, variance and standard deviation measure how spread out your dataset is.0257

In this case, since we got a uniform distribution, if it is spread out over a wider area0264

then you should have a higher variance or a higher standard deviation.0270

If it is squished into a smaller area then you should have a smaller variance or a smaller standard deviation.0274

In this case, since we have the term of θ2 – θ1, that is the width of the interval there, θ2 – tθ1.0282

What we are saying here is that the standard deviation is proportional to the width of the interval.0291

If you have a wider interval, if you spread your data out more then you have a larger standard deviation.0297

If you compress your interval, then you will have a smaller standard deviation.0303

It should be intuitively plausible.0307

What is not so obvious I think is, the constant 12 for the variance, 2√ 3 for the standard deviation.0310

I think that part would not be obvious unless, you actually calculated them out.0317

Let us go ahead and look at some problems involving the uniform distribution.0322

They generally tend to be fairly easy to calculate.0328

The first example here is, you are sitting on your front doorstep, waiting for your morning newspaper to arrive.0330

It always arrives sometime between 7:00 AM and the noon.0337

The time at which it arrives follows a uniform distribution.0342

We want to find that probability that it will arrive during an odd number hour,0347

which means we want it to arrive between 7 and 8, not between 8 and 9 because that would be even number hour.0352

Between 9 and 10 we are qualified, 10 and 11 would not qualify, and 11 to 12 that would also qualify as an odd numbered hour.0360

It looks like there is 3 hours here that would qualify.0371

Our total range here is 7 to noon, 7 to 12, that is θ1 is 7 and θ2 is 12.0374

Θ2- θ1 is 12 – 7 is 5, that is our denominator here.0385

We want to talk about what range we are interested in.0401

We have all these odd number hours, 8 -7 + 10 - 9 + 12 -11, which of course is just 3 separate hours on that 5 hour interval.0404

We have a total probability of 3/5.0423

If your newspaper is going to arrive sometime between 7:00 AM and 12, and it is uniformly distributed in that interval,0431

then there is a 60% chance that it will be an odd numbered hour.0440

Let me recap that.0447

We got this newspaper arriving in a 5 hour interval, that is where I got that denominator of 5 because it is a 5 hour interval.0449

If you want think about it in terms of θ2 – θ1, that is 12 -7, that is where that 5 comes from.0457

The 3 comes from the 3 hours that have odd numbers.0463

The 7:00 hour, the 9:00 hour, and 11:00 hour, gives you 3 different hours that have odd numbers.0469

Our total fraction is 3/5, the probability that it will arrive during an odd numbered hour is exactly 60%.0477

Remember, I told you that the uniform distribution is one of the easiest to deal with.0484

Problems involving uniform distribution are often very easy computationally.0489

This certainly qualifies as an easy one computationally but if you stick around, I got a couple of harder once coming up.0494

We will see something a little harder coming up.0500

Example 2 is also going to be an easy one, trust me, example 3 will be a little more tricky.0504

But example 2, we are going to pick a real number Y from the uniform distribution on the interval from 5 to 12.0511

Let me go ahead and graph this out.0519

There is 5, there is 6, 7, 8, 9, 10, 11, there is 12.0521

We are going to pick a real number Y from somewhere in this interval on the uniform distribution.0531

We want to find the probability that that value of Y will turn out to be bigger than 9.0537

The probability that Y would be bigger than or equal to 9, is equal to, on the denominator we have θ2 – θ1.0544

In the numerator, we have B - A being the interval that we are interested in.0555

Let me go ahead and draw that in.0561

Θ1 is 5, θ2 is 12, the interval that we are interested in is from 9 to 12 because we want Y to be bigger than 9.0563

6, 7, 8, 9, there is 9 right there.0573

There is A is 9, and B is the same as θ2, B is 12.0577

B - A is 12 – 9, θ2 – θ1 is 12 -5.0583

What we have there is 12 – 9 is 3, 12 -5 is 7.0592

The probability that our number will be bigger than 9 is exactly 3/7.0598

Again, very easy computations for the uniform distribution, it is just a matter of saying how wide is your interval.0604

In this case, our interval is 7 units wide, the interval from 5 to 12, that is where we got that 7 in the denominator.0612

How wide is the region that you are interested in, the region that you might call success?0621

In this case, success is defined as Y is bigger than 9.0626

That region of success will be the interval from 9 to 12.0631

The width of that interval from 9 to 12 is just 12 - 9 is 3, that is where we got that 3.0637

Our total answer, our total probability there is just 3/7.0643

I guess you could convert that into a decimal, I think that would turn out to be about 42%.0649

But that does not come out neatly, I’m going to leave it as a fraction as 3/7 there.0655

Example 3 is a bit trickier, what is happening here is that you have a dinner date with your best friend.0663

You are planning to meet at 6 pm at the restaurant but the problem is that, both of you tend to run a little late.0670

In fact, even though you are planning to me at 6:00 pm, you might be a little bit late, your friend might be a little bit late.0678

One of you is probably going to end up waiting a little bit for the other one.0687

The way it works is you tend to arrive between 0 and 15 minutes late.0690

You are never later than 15 minutes and you are never early.0696

You are always maybe 7 minutes, 8 minutes, 11 minutes late, somewhere between 0 and 15.0700

Your friend is always between 0 and 10 minutes late.0706

Your friend is a little bit more prompt than you are.0709

The question we are asking here is the probability that you will arrive before your friend.0713

In other words, will you be the one who gets there first and has to find the table at dinner0718

and will be sitting around waiting for your friend, or will it be the other way around?0724

Your friend arrives first and has to deal with the waiter, and your friend would be waiting for you.0728

Let me show you how to solve this one.0733

We are going to set up two variables here because there are two independent things going on.0736

There is you arriving to the restaurant and there is your friend arriving to the restaurant.0740

I set up the variable for X which is your arrival time which could be anywhere from 0 to 15 minutes late.0746

Y is going to be your friends arrival time which could be anywhere from 0 to 10 minutes late.0759

A really useful way to think about this problem is to graph it.0770

Let me go ahead and draw a graph of the possibilities here.0774

I will put X on the X axis and Y on the Y axis.0780

Your arrival time can be anywhere from 0 to, there is you being 5 minutes late, here you are being 10 minutes late,0785

and here you are coming in 15 minutes late.0794

We know you are not going to be later than that.0797

There is your friend arriving 5 minutes late and here is your friend arriving 10 minutes late.0800

We know that your friend would not be later than 10 minutes.0806

What that means is your combined arrival time, the combination of arrival times0809

is going to be somewhere in this rectangle, depending on when you arrive and depending on when your friend arrives.0817

Somewhere, you will arrive a certain number of minutes late and your friend will arrive a certain number of minutes late.0825

That will give us a point somewhere in this rectangle.0830

Then, we will look at that and say did you arrive first or did your friend arrive first.0834

The way we want to think about that is we want to calculate the probability that you will arrive before your friend.0841

In other words, we want the probability that X is less than or equal to Y, that is you arriving before your friend.0848

Let me just turn those variables around because I think it will be easier to graph that way.0859

The probability that Y is greater than or equal to X, that is saying the same thing.0864

Let me graph the region in which Y is greater than X.0869

A little bit of algebra review here.0873

Maybe, I will do this in black.0877

The line Y equals X is that line right there.0881

That is the line Y is equal to X.0885

Y greater than X means you go above the lines.0888

It is all this triangular region above the line here.0894

That is the region where you arrive before your friend.0900

Anywhere below the line, that means your friend arrives first and you arrived afterwards.0907

Let us try to calculate that the probability of being in that black shaded region.0915

It is the total shaded area ÷ the total area in the rectangle.0921

Let us try to figure out what those areas are.0932

The shaded area, I see I have a triangle with 10 units on a side here.0934

That is base × height/2, that is 10 × 10/2, 100/2 is 50.0942

50 units in your shaded triangle.0950

The total area is a rectangle, it is 10 by 15 on the side, that is 10 × 15 is 150.0955

Here is very nice, it simplifies lovely to 1/3.0963

The probability that you will arrive before your friend is exactly 1/3.0970

If you make lots and lots of dates with the same friend, and you guys both follow the same habits over the years,0977

what will happen is 1/3 of the time you will be sitting around waiting for your friend.0985

2/3 of the time your friend will be sitting around waiting for you.0990

That is really the result of the fact that your friend is a little bit more prompt than you.0995

Most of the time your friend will end up waiting for you, at some of the time, 1/3 of the time,1003

you will wait for your friend.1007

Let me recap that.1009

The way we approach this is, we noticed first that we really had two independent uniform distributions.1011

There is one for your arrival time and there is one for your friend’s arrival time.1017

That is the first thing I did was to set up variables to indicate your arrival time and your friend’s arrival time.1023

Your arrival time, I put on the X axis that goes from 0 to 15 because you can be anywhere from 0 to 15 minutes late.1030

Your friend is anywhere from 0 to 10 minutes late.1040

I got those from the stem of the problem here.1043

I graph those here and I got this nice rectangle of possible combinations of arrival ×.1048

Once I have this rectangle, I know that while you arrive at a particular time, your friend will arrive at a particular time,1057

that means essentially we are choosing a point at random in this rectangle.1063

And then, we have to ask whether you will arrive before your friend?1069

You arriving before your friend means your arrival time is less than or equal to your friend’s arrival time,1074

which can be rewritten as Y greater than X.1081

We graph the shaded region, represents the region where Y is greater than or equal to X.1084

And then, it was just a matter of calculating the areas which turned into1091

a little old fashion geometry of calculating the area of a triangle.1094

The triangle was ½ base × height, ½ × 10 × 10.1099

The rectangle was base × height, that is 15 × 10.1105

We get 50/150 simplifies down to a probability of 1/3.1113

That represents the chance that you have arrived at the restaurant before your friend.1119

You will be the one who has to sit around and wait.1125

In example 4, we have a problem that is a great interest to computer programmers.1132

The reason is that most computer languages have a random number function.1139

It uses something rand or random, or something like that.1146

If you type rand into a computer program and the appropriate language, it will give you a number between 0 and 1.1151

We usually try to arrange it so that the random numbers are uniformly distributed between 0 and 1,1162

which is very good if you need a number between 0 and 1.1168

In lots of cases, when you need a random number in a computer program,1172

you need a random number on another range which might be from θ1 to θ2.1177

What this problem really does is, it shows you how to convert a uniform distribution on 0-11184

into a uniform distribution onto θ1 and θ2.1193

That is the point of this problem, it is very useful for computer programmers1197

but that is not actually what we are doing here.1203

What we are doing is we are making a little transformation and we are starting with Y1205

being uniformly distributed on the interval from 0 to 1.1210

We are looking at another variable X which is defined to be, by the way that colon means define to be.1215

X is defined to be θ2 – tθ1 Y + θ1.1230

We want to show that that is a uniform distribution on the interval from θ1 to θ2.1236

To show that, let me first find the range of values for X.1243

Notice that, the range for Y is from 0 to 1, I have Y =0 to Y =1, if I plug those values into X,1250

if I plug Y =0 in to X, I get X = θ2 – θ1 × 0, that drops out.1260

I just get X = θ1.1268

If I plug Y = 1 into X, I get X = θ2- θ1 × 1 + θ1.1270

That simplifies down to θ2.1282

That tells me the range for X, X goes from θ1 to θ2.1287

That is hopeful, at least I know that X is distributed somehow on the range from θ1 to θ2.1294

But I want to really show that X is a uniform distribution.1302

I want to calculate the probability of the line between any 2 values.1305

Let me find that probability here.1313

The probability that is X is between any two values A and B, I can calculate that as the probability that,1317

X, just by definition is θ2 – θ1 × Y + θ1, that should be between A and B.1330

What I want to do is to solve this into a set of probabilities for Y.1343

First, subtract off θ1 from all 3 sides there.1350

The probability of A – θ1 being less than or equal to θ2 -θ1 × Y, less than or equal to B – θ1.1354

I’m trying to solve for Y, I'm trying to get Y by itself.1368

Next, I’m going to divide by θ2 – θ1.1372

This is the probability of A – θ1/θ2- θ1.1377

Less than or equal to just Y by itself now, less than or equal to B –θ1/θ2 – θ1.1385

I'm remembering that Y itself is a uniformly distributed random variable.1397

The probability that Y would be between any two bounds is just the difference between those two bounds.1404

You just subtract those two bounds.1411

I will just do B – θ1/θ2- θ1- A – θ1/θ2 – θ1.1413

I see now that I have a common denominator, θ2- θ1.1425

I got –θ1 in the first term and - and -, + θ1 in the second term.1434

Those θ1 cancel with each other and I just get down to B – A.1441

If you look at that, what I did was I started out with the probability that X is going to be between A and B.1448

What I came up with is, the probability is equal to exactly B - A ÷ θ2 – θ1.1458

That is exactly the formula for a uniform distribution.1468

X has a uniform distribution, X is uniformly distributed, distribution on the interval from θ1 to θ2.1475

I should start with θ1 and go to θ2.1501

X has a uniform distribution because the probability of X falling in any interval is exactly equal to the width of that interval B – A.1507

To recap what I did there, first, I looked at the range for Y, Y goes from 0 to 1.1519

Based on that, I calculated the range for X.1526

I plugged in those values of Y 0 and 1 into the formula for X here, and calculated the bounds for X being θ1 to θ2.1529

I know that X takes on the right range of values.1541

And then, I found the probability of any particular sub interval from A to B by converting X into terms of Y.1545

Solving out to isolate the variable Y, and then I use the fact that Y is uniformly distributed.1556

The probability of Y being between any two limits is just the width of those limits, the difference between those two limits.1563

B – θ1/θ2 –θ1- A – θ1/θ2 – θ1.1572

That simplify down to B –A /θ2 – θ1, which is exactly the formula for probability with a uniform distribution.1580

That tells me that X has a uniform distribution on the interval from θ1 to θ2.1594

That is very useful if you are computer programmer because that means1601

you can use the random number generator given by most computer programming languages.1605

And then, you can use this formula to convert it into a uniform distribution whatever range you want.1613

If you want for example, a random number between 80 and 100, and then you just plug in θ1 = 80 and θ2 =100.1620

You can use this formula to generate a random number between 80 and 100,1631

that will be uniformly distributed between 80 and 100.1638

In example 5, we are going leave the world of the random numbers behind.1644

We are going to look at the rough and tumble world of ice cream dispensary.1649

We have an ice cream machine which gives you servings of ice cream.1653

The servings vary a little bit, if you are unlucky the machine will be stingy with you, and it will give you just 206 ml of ice cream.1659

If it is a good day for you, if the machine is feeling generous, it will give you up to 238 ml.1668

Essentially, it picks a random amount in between 206 and 230, and they are uniformly distributed.1676

The question we are trying to answer is, if you go up there with your bowl and1684

you want to predict how much ice cream you will get, you want to describe what the expected amount of ice cream is in a serving,1691

and also the standard deviation in that quantity.1698

This is really asking, the expected value and mean are the same thing.1705

We are trying to calculate the expected value or the mean of this uniform distribution, and also the standard deviation.1710

I gave you formulas for those as few slides back, in a slide called key properties of the uniform distribution.1717

You can go back and look those up, I will remind you what they are here.1728

The mean which is always the same as the expected value, by definition those are the same thing.1732

For the uniform distribution is θ1 + tθ2 ÷ 2.1738

The θ1 and θ2 are the ranges of the endpoints of the interval.1746

In this case that is 206 + 230 ÷ 2, that is 436 ÷ 2 which is 218.1752

Our units here are ml, the average amount you expect to get when you fill up your bowl1767

at this ice cream machine is 218 ml of ice cream.1775

Of course that is not at all surprising, if you are going to get a random amount between 206 and 230,1780

it is not surprising that in the long run, you will get about halfway between 206 and 230 which is 218.1789

That is really not surprising at all.1798

The standard deviation, I also gave you on that slide, several slides ago.1801

It is θ2- θ1 ÷ 2 √s 3.1805

In this case, θ2, the big one is 230, Θ1 is 206, we want to divide that by 2 √3.1812

230 -206 is 24, the 200 part cancels, ÷ 2 and 3, that simplifies down to 12/3.1823

Since, 12 is 4 × 3, this just gives us 4 × √ 3 which I put that into a calculator,1837

it works out to be just a little bit less than 7 ml.1849

It is about 6.9 ml.1854

If you fill up your bowl at this ice cream machine, you expect on average to get about 218 ml.1859

The standard deviation on that will be 6.9 ml, about 7 ml + or - from 200 and 18.1867

To recap where these numbers came from.1876

The formulas for the mean and standard deviation, I give this to you on early slide in this talk.1879

It was called key properties of the uniform distribution.1885

They are fairly straightforward formulas.1888

In particular, the mean is what you would guess.1889

It is just the average of the upper and lower bounds, 206 + 230/2 gives you an average of 280 ml of ice cream per serving.1893

The standard deviation is probably not something you would guess but if you have a formula handy, it is θ2 – θ1/203.1905

I will just plug in the θ2 and θ1 into those values there.1914

And I simplified it down to 4 √3 and I have this decimal approximation that is about 6.9 or about 7 ml.1920

That is your standard deviation in an ice cream serving.1928

That is the last example and that wraps up this lecture on the uniform distribution.1932

The uniform distribution is just the first, it is the easiest of several continuous distributions.1938

We will be moving on from here and looking at the famous normal distribution, not the same as the uniform distribution,1944

and also the gamma distribution which includes the exponential distribution and chi square distribution.1951

Those are all coming up in the next few lectures here in the probabilities series on www.educator.com.1958

You are watching Will Murray with www.educator.com, and thank you very much for joining us, bye.1965